Problem 6 - Master Resource Book in JEE Main Physics

Question

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with \(1 \mathrm{MeV}\) energy is nearly (a) \(1.2 \mathrm{~nm}\) (b) \(1.2 \times 10^{-3} \mathrm{~nm}\) (c) \(1.2 \times 10^{-6} \mathrm{~nm}\) (d) \(1.2 \times 10^{1} \mathrm{~nm}\)

Step-by-Step Solution

Answer

The wavelength is nearly \(1.2 \times 10^{-3} \text{ nm}\), so option (b) is correct.

Step 1: Convert Energy from MeV to Joules

The energy required to remove the proton is given in mega electronvolts (MeV). We first need to convert this energy to joules (J). The conversion factor is: \[ 1 ext{ MeV} = 1.602 imes 10^{-13} ext{ J} \]Given the energy is 1 MeV, we multiply by the conversion factor to find the energy in joules:\[ E = 1 ext{ MeV} \times 1.602 imes 10^{-13} ext{ J/MeV} = 1.602 imes 10^{-13} ext{ J} \]

Step 2: Use the Energy-Wavelength Relationship

The energy of a photon is related to its wavelength by the equation:\[ E = \frac{hc}{\lambda} \]Where:- \( E \) is the energy in joules, which we calculated as \( 1.602 \times 10^{-13} \text{ J} \)- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \text{ J s} \)- \( c \) is the speed of light, approximately \( 3.00 \times 10^{8} \text{ m/s} \)- \( \lambda \) is the wavelength, which we need to find.

Step 3: Solve for Wavelength

Rearrange the equation to solve for wavelength \( \lambda \):\[ \lambda = \frac{hc}{E} \]Substitute the known values into the equation:\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.00 \times 10^{8} \text{ m/s}}{1.602 \times 10^{-13} \text{ J}} \]Calculate to find:\[ \lambda \approx 1.24 \times 10^{-12} \text{ m} \]

Step 4: Convert Wavelength from Meters to Nanometers

Convert the wavelength from meters to nanometers by using the conversion factor:\[ 1 ext{ m} = 1 imes 10^{9} ext{ nm} \]Thus:\[ \lambda \approx 1.24 \times 10^{-12} ext{ m} \times 1 imes 10^{9} ext{ nm/m} \approx 1.24 \times 10^{-3} ext{ nm} \]

Key Concepts

Energy Conversion

When dealing with problems in quantum physics, energy conversion is a fundamental step. We often encounter different units of energy, like mega electronvolts (MeV) and joules (J). The relationship between these units is vital for converting energy into a usable form for calculations.
In many quantum calculations, energy is initially given in electronvolts (eV), which is a convenient unit for expressing the energy of particles such as electrons or photons. If the energy is given in mega electronvolts (MeV), it means a million electronvolts. To use this quantity in formulas that involve joules, a conversion is necessary.

• The conversion factor to remember is: \( 1 \, \text{MeV} = 1.602 \times 10^{-13} \, \text{J} \).

Using this conversion, we can translate the given energy into joules before further solving, ensuring consistent units throughout the calculation.

Energy-Wavelength Relationship

The energy-wavelength relationship is a cornerstone concept in understanding photon behavior. This relationship is articulated through the equation:\[ E = \frac{hc}{\lambda} \]Here, \( E \) is the photon's energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. This equation shows how photons with higher energies correspond to shorter wavelengths, and vice versa.
In practice, once energy has been converted into joules, this formula enables you to solve for the wavelength. You rearrange the formula to isolate \( \lambda \):
\[ \lambda = \frac{hc}{E} \]This simple manipulation demonstrates the inverse relationship between energy and wavelength. Knowing the constants \( h \) and \( c \) allows you to easily compute the wavelength given the energy, bridging the gap between the two properties.

Quantum Physics

Quantum physics provides deep insight into the behavior of particles on the smallest scales, where traditional physics often fails to apply. One key aspect is the photon, a fundamental particle in quantum mechanics that exhibits both wave and particle properties.
Photons carry energy and are fundamentally linked to electromagnetic radiation. Although massless, they possess momentum and can interact with particles, such as causing the emission of electrons from metals in the photoelectric effect. Understanding photons' behavior requires grasping both their energy and wavelength.

• This includes knowing how to calculate energy using the Planck-Einstein relation and understand their wavelength via the energy-wavelength relationship.

These concepts highlight the quantum view where different particles are described by both particle-like and wave-like characteristics, giving a comprehensive understanding of phenomena that classical physics can't explain.